Basic Algebra 2

$\color{#3D99F6}{\frac{\sqrt{a^2-a\sqrt{12}+3}}{a\sqrt{3}-3}=\frac{\sqrt{a^2-2a\sqrt{3}+(\sqrt{3})^2}}{a\sqrt{3}-3}}$ $\color{#3D99F6}{\frac{\sqrt{(a-\sqrt{3})^2}}{a\sqrt{3}-3}=\frac{a-\sqrt{3}}{a\sqrt{3}-3}}$ $\color{#3D99F6}{\frac{(a-\sqrt{3})(a\sqrt{3}+3)}{(a\sqrt{3}-3)(a\sqrt{3}+3)}}$ $\color{#3D99F6}{\frac{a^2\sqrt{3}+3a-3a-3\sqrt{3}}{3a^2-9}}$ $\color{#3D99F6}{\frac{a^2\sqrt{3}-3\sqrt{3}}{3a^2-9}=\frac{\sqrt{3}(a^2-3)}{3(a^2-3)}=\frac{\sqrt{3}}{3}}$ I got the answer like this without the negative sign but I proceeded with my solution anyways.So $\color{#D61F06}{s=3}$ and $s^s=3^3=\boxed{27}$

keeping in mind a<1 numerator of given can be written as
[{(3^1/2)-a}^2]^(1/2)=3^1/2-a= -(a-3^1/2) and
denominator can be written as (3^1/2)(a-3^1/2)
now nume./deno.= -1/(3^1/2)=-(3^1/2)/3
Hence answer is 3^3=27

In the numerator , a={√12±√(√12-4*3)}/2=√3 a^{2}-a√12+3=(a-√3)^{2} \frac{(a^{2}-a√12+3)^{1/2}}{a√3-3}=\frac{((a-√3)^{2})^{1/2}}{√3(a-√3)} =\frac{1}{√3}=\frac{√3}{3} Therefore s=3. s^{s}=3^{3}=27